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Stels [109]
3 years ago
6

le" class="latex-formula">
​

Mathematics
1 answer:
Zina [86]3 years ago
3 0

Answer:

\large\boxed{19\dfrac{17}{20}}

Step-by-step explanation:

\text{First step:}\\\\\text{Find LCD}\ (Least\ Common\ Denominator)\\\\LCD=20\\\\4\cdot5=20\\5\cdot4=20\\\\\dfrac{1}{4}=\dfrac{1\cdot5}{4\cdot5}=\dfrac{5}{20}\\\\\dfrac{3}{5}=\dfrac{3\cdot4}{5\cdot4}=\dfrac{12}{20}\\\\9\dfrac{1}{4}+10\dfrac{3}{5}=9\dfrac{5}{20}+10\dfrac{12}{20}=(9+10)+\dfrac{5+12}{20}=19\dfrac{17}{20}

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An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t
USPshnik [31]
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

\displaystyle\int_a^bf(x)\,\mathrm dx

Split up the interval [a,b] into n equal subintervals,

[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]

where a=x_0 and b=x_n. Each subinterval has measure (width) \dfrac{a-b}n.

Now denote the left- and right-endpoint approximations by L and R, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are \{x_0,x_1,\cdots,x_{n-1}\}. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, \{x_1,x_2,\cdots,x_n\}.

So, you have

L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)
R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)

Now let T denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)

Factoring out \dfrac12 and regrouping the terms, you have

T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)

which is equivalent to

T=\dfrac12\left(L+R)

and is the average of L and R.

So the trapezoidal approximation for your problem should be \dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2
4 0
3 years ago
Write the equation of the line in point-slope form that goes through the points (0, 1) and (2, 7)
laiz [17]

Answer:

point - slope form

 y - 1 = 3 (x-0)

Step-by-step explanation:

<u><em>step(i):-</em></u>

Given points are (0,1) and (2,7)

The slope of the line

     m = \frac{y_{2}-y_{1}  }{x_{2} -x_{1} }

    m = \frac{y_{2}-y_{1}  }{x_{2} -x_{1} } = \frac{7-1}{2-0} = \frac{6}{2}  = 3

  m =3

<u><em>Step(ii):-</em></u>

point -slope form

   y - y₁ = m(x-x₁)

Equation of the straight line passing through the point (0,1) and having slope 'm' = 3

   y - 1 = 3 (x-0)

  y -1 = 3x

 3x - y +1=0

Equation of the straight line passing through the point (0,1) and having slope 'm' = 3 is 3x -y +1=0

4 0
2 years ago
Can someone please help me with this problem
Nadusha1986 [10]
X= 2/3

6x/x-6 - 4/x - 24 / x^2 - 6x = 0


6x^2 -4 (x-6)-24 / x(x-6) = 0

6x^2 -4x+ 24 -24/x(x-6) = 0

X(6x-4) / x(x-6)

6x-4/x-5 = 0

6x-4= 0

6x = 4 divide both sides by 6
X= 2/3
5 0
3 years ago
A triangle has one side which 2 meters long and another side which is 5 meters long write a value that could be the length of th
Firlakuza [10]

Answer:

The measure of the third side should be less than 2+5 = 7

Length of the third side<7

Hope this helps!

7 0
3 years ago
All of the following expressions are equivalent to 12w + 6, except:
laiz [17]
-1(12w - 6) is not equivalent because when the -1 is distributed, the equation becomes -12w + 6, not 12w + 6
5 0
3 years ago
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